Q.Integrate the following function: $\frac{x+2}{\sqrt{4x - x^2}}$

Complete the square in the denominator and rewrite the numerator to match the derivative of the radicand.

Write $x+2 = -\tfrac12(4-2x) + 4$ (check: $-2+x+4 = x+2$). Then

First integral: with $u = 4x-x^2$, $du=(4-2x)dx$, it is $-\tfrac12\cdot 2\sqrt{u} = -\sqrt{4x-x^2}.$ …

Setting up

A square root of a quadratic invites completing the square:

So $\sqrt{4x-x^2} = \sqrt{4-(x-2)^2}$, the classic $\sqrt{a^2-u^2}$ shape with $a=2$.

The numerator split

The derivative of the radicand is $\dfrac{d}{dx}(4x-x^2) = 4-2x$. We peel that off the numerator: solve $x+2 = A(4-2x)+B$. Matching $x$: $1=-2A\Rightarrow A=-\tfrac12$; matching constants: $2 = 4A+B = -2+B \Rightarrow B=4$. Hence

Step 1 — the derivative part

Let $u = 4x-x^2$, so $du = (4-2x)\,dx$:

Step 2 — the arcsine part …

Method: $\dfrac{\text{linear}}{\sqrt{\text{quadratic}}}$ with a negative $x^2$ term

Same split as any linear-over-root problem, but because the $x^2$ coefficient is negative the constant piece becomes an inverse sine, and the derivative of the radicand carries a leading minus sign.

Steps

Step 1: Split the numerator. For $Q(x)=c+bx-x^2$, note $Q'(x)=b-2x$; write the numerator as $\lambda(b-2x)+\mu$.

Step 2: Derivative part. …

Common Mistakes

Mistake 1: Getting the sign of $\lambda$ wrong.

Why it's wrong: the derivative of $4x-x^2$ is $4-2x$, so $x+2=-\tfrac12(4-2x)+4$; missing the minus flips the $\sqrt{Q}$ term's sign. Correct approach: differentiate the radicand exactly (with its minus) before matching.

Mistake 2: Using a log form for the constant part.

Why it's wrong: the negative $x^2$ term gives $a^2-u^2$, hence $\sin^{-1}$. Correct approach: complete the square and check you have $a^2-u^2$, not $u^2\pm a^2$. …